Can I use PNT in this way? I want to show that for large $n$ the $n$-th prime grows like $n\ln (n)$. Is this correct?
By PNT
$$\mathop {\lim }\limits_{x \to \infty } \frac{{\pi (x)\ln (x)}}{x} = 1.$$
Let $x = {p_n}$, so that $\pi (x) = n$, where ${p_n}$ is the $n$-th prime, then
$$\mathop {\lim }\limits_{n \to \infty } \frac{{n\ln (n)}}{{{p_n}}} = 1$$
or
$${p_n} \sim n\ln (n).$$
 A: To complete your argument you have to show that $\ln(p_n)/\ln(n) = 1.$  
Now, if you have that $p_n \sim n\ln (n),$ then $\ln(p_n) = \ln(n) + \ln\ln(n) + o(1)$, and so indeed $\ln(p_n)/\ln(n) \to 1$ as $n \to \infty$.  So at least
this is consistent with what you are trying to prove.
On the other other hand, what you know is that
$n\ln(p_n)/p_n \to 1$ as $n \to \infty$.  Taking $\ln$ gives
$$\ln(n) + \ln\ln(p_n) = \ln(p_n) + o(1),$$
and so
$$\ln(n)/\ln(p_n) + \ln\ln(p_n)/\ln(p_n) = 1 + o(1).$$
Can you finish from there?
A: There is a problem with this approach. You took $x = p_n$, but that means $\ln(x) = \ln(p_n)$ and not $\ln(n)$.
A: Assume $(1):\hspace{3mm}\frac{\pi(x) \log x}{x}\sim 1 $, take logs of both sides: 
$\lim_{x \to \infty}(\log \pi(x)+ \log\log x - \log x) = 0$
or $$ \lim_{n \to \infty} [ \log x(\frac{\log \pi(x)}{\log x} + \frac{\log\log x}{\log x} -1)] = 0$$ 
Because $\log x \to \infty$ as $ x \to \infty,$
$$\lim (\frac{\log \pi(x)}{\log x} + \frac{\log\log x}{\log x} -1) = 0 $$
So $$(2) \hspace{3mm}\lim_{x\to\infty} \frac{\log \pi(x)}{\log x} = 1$$ and from (1)  we have 
$$(3)\hspace{3mm}\lim \frac{\pi(x) \log \pi(x)}{x} = 1 $$ and if $x = p_n, \pi(x) = n, $ so $\pi(x) \log \pi(x) = n \log n$, so (3) implies that 
$$\lim_{n\to \infty }\frac{n \log n}{p_n } = 1.  $$
This is basically Apostol's proof. 
